Construction of Longest Increasing Subsequence (N log N)

In my previous post, I have explained about longest increasing sub-sequence (LIS) problem in detail. However, the post only covered code related to querying size of LIS, but not the construction of LIS. I left it as an exercise. If you have solved, cheers. If not, you are not alone, here is code.

If you have not read my previous post, read here. Note that the below code prints LIS in reverse order. We can modify print order using a stack (explicit or system stack). I am leaving explanation as an exercise (easy).

CPP

// C++ implementation to find longest increasing subsequence
// in O(n Log n) time.
#include <bits/stdc++.h>
using namespace std;

// Binary search
int GetCeilIndex(int arr[], vector<int> &T, int l, int r,
                 int key)
{
    while (r - l > 1)
    {
        int m = l + (r - l)/2;
        if (arr[T[m]] >= key)
            r = m;
        else
            l = m;
    }

    return r;
}

int LongestIncreasingSubsequence(int arr[], int n)
{
    // Add boundary case, when array n is zero
    // Depend on smart pointers

    vector<int> tailIndices(n, 0); // Initialized with 0 
    vector<int> prevIndices(n, -1); // initialized with -1

    int len = 1; // it will always point to empty location
    for (int i = 1; i < n; i++)
    {
        if (arr[i] < arr[tailIndices[0]])
        {
            // new smallest value
            tailIndices[0] = i;
        }
        else if (arr[i] > arr[tailIndices[len-1]])
        {
            // arr[i] wants to extend largest subsequence
            prevIndices[i] = tailIndices[len-1];
            tailIndices[len++] = i;
        }
        else
        {
            // arr[i] wants to be a potential condidate of
            // future subsequence
            // It will replace ceil value in tailIndices
            int pos = GetCeilIndex(arr, tailIndices, -1,
                                   len-1, arr[i]);

            prevIndices[i] = tailIndices[pos-1];
            tailIndices[pos] = i;
        }
    }

    cout << "LIS of given input" << endl;
    for (int i = tailIndices[len-1]; i >= 0; i = prevIndices[i])
        cout << arr[i] << " ";
    cout << endl;

    return len;
}

int main()
{
    int arr[] = { 2, 5, 3, 7, 11, 8, 10, 13, 6 };
    int n = sizeof(arr)/sizeof(arr[0]);

    printf("LIS size %d\n", LongestIncreasingSubsequence(arr, n));

    return 0;
}

Java

// Java implementation to find longest 
// increasing subsequence in O(n Log n)
// time.
import java.util.Arrays;

class GFG {
    
    // Binary search
    static int GetCeilIndex(int arr[], 
                       int T[], int l, 
                       int r, int key)
    {
        
        while (r - l > 1) {
            
            int m = l + (r - l) / 2;
            if (arr[T[m]] >= key)
                r = m;
            else
                l = m;
        }

        return r;
    }

    static int LongestIncreasingSubsequence(
                           int arr[], int n)
    {
        
        // Add boundary case, when array n is zero
        // Depend on smart pointers

        int tailIndices[] = new int[n];
        
        // Initialized with 0
        Arrays.fill(tailIndices, 0); 
        
        int prevIndices[] = new int[n];
        
        // initialized with -1
        Arrays.fill(prevIndices, -1); 
        
        // it will always point to empty 
        // location
        int len = 1; 
        
        for (int i = 1; i < n; i++) {
            if (arr[i] < arr[tailIndices[0]]) 

                // new smallest value
                tailIndices[0] = i;
            
            else if (arr[i] > 
                arr[tailIndices[len - 1]]) {
                    
                // arr[i] wants to extend 
                // largest subsequence
                prevIndices[i] = tailIndices[len - 1];
                tailIndices[len++] = i;
            }
            else {

                // arr[i] wants to be a potential 
                // condidate of future subsequence
                // It will replace ceil value in 
                // tailIndices
                int pos = GetCeilIndex(arr, 
                    tailIndices, -1, len - 1, arr[i]);

                prevIndices[i] = tailIndices[pos - 1];
                tailIndices[pos] = i;
            }
        }

        System.out.println("LIS of given input");
        
        for (int i = tailIndices[len - 1]; i >= 0; 
                                i = prevIndices[i])
            System.out.print(arr[i] + " ");
            
        System.out.println();

        return len;
    }
    
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 5, 3, 7, 11, 8, 10, 13, 6 };
        int n = arr.length;

        System.out.print("LIS size\n" + 
                LongestIncreasingSubsequence(arr, n));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python implementation to
# find longest increasing
# subsequence
# in O(n Log n) time.

# Binary search
def GetCeilIndex(arr, T, l, r, key):

    while (r - l > 1):
    
        m = l + (r - l)//2
        if (arr[T[m]] >= key):
            r = m
        else:
            l = m

    return r
 
def LongestIncreasingSubsequence(arr,n):

    # Add boundary case,
    # when array n is zero
    # Depend on smart pointers
    
    # Initialized with 0
    tailIndices=[0 for i in range(n+1)]  

    # Initialized with -1
    prevIndices=[-1 for i in range(n+1)]  
    
    # it will always point
    # to empty location
    len = 1 
    for i in range(1, n):
    
        if (arr[i] < arr[tailIndices[0]]):
        
            # new smallest value
            tailIndices[0] = i
        
        elif (arr[i] > arr[tailIndices[len-1]]):
        
            # arr[i] wants to extend
            # largest subsequence
            prevIndices[i] = tailIndices[len-1]
            tailIndices[len] = i
            len += 1
        
        else:
        
            # arr[i] wants to be a
            # potential condidate of
            # future subsequence
            # It will replace ceil
            # value in tailIndices
            pos = GetCeilIndex(arr, tailIndices, -1,
                                   len-1, arr[i])
 
            prevIndices[i] = tailIndices[pos-1]
            tailIndices[pos] = i
        
    print("LIS of given input")
    i = tailIndices[len-1]
    while(i >= 0):
        print(arr[i] , " ",end="")
        i = prevIndices[i]
    print()
 
    return len

# driver code
arr = [ 2, 5, 3, 7, 11, 8, 10, 13, 6 ]
n = len(arr)
 
print("LIS size\n",LongestIncreasingSubsequence(arr, n))

# This code is contributed
# by Anant Agarwal.

C#

// C# implementation to find longest 
// increasing subsequence in O(n Log n)
// time.
using System;

class GFG {
    
    // Binary search
    static int GetCeilIndex(int []arr, int []T, int l, 
                                       int r, int key)
    {
        
        while (r - l > 1)
        {
            int m = l + (r - l) / 2;
            
            if (arr[T[m]] >= key)
                r = m;
            else
                l = m;
        }

        return r;
    }

    static int LongestIncreasingSubsequence(
                                   int []arr, int n)
    {
        
        // Add boundary case, when array n is zero
        // Depend on smart pointers

        int []tailIndices = new int[n];
        
        // Initialized with 0
        for(int i = 0; i < n; i++)
            tailIndices[i] = 0; 
    
        
        int []prevIndices = new int[n];
        
        // initialized with -1
        for(int i = 0; i < n; i++)
            prevIndices[i] = -1;
        
        
        // it will always point to empty 
        // location
        int len = 1; 
        
        for (int i = 1; i < n; i++) {
            if (arr[i] < arr[tailIndices[0]]) 

                // new smallest value
                tailIndices[0] = i;
            
            else if (arr[i] > 
                arr[tailIndices[len - 1]]) {
                    
                // arr[i] wants to extend 
                // largest subsequence
                prevIndices[i] = tailIndices[len - 1];
                tailIndices[len++] = i;
            }
            else {

                // arr[i] wants to be a potential 
                // condidate of future subsequence
                // It will replace ceil value in 
                // tailIndices
                int pos = GetCeilIndex(arr, 
                    tailIndices, -1, len - 1, arr[i]);

                prevIndices[i] = tailIndices[pos - 1];
                tailIndices[pos] = i;
            }
        }

        Console.Write("LIS of given input");
        
        for (int i = tailIndices[len - 1]; i >= 0; 
                                i = prevIndices[i])
            Console.Write(arr[i] + " ");
            
        Console.WriteLine();

        return len;
    }
    
    // Driver code
    public static void Main()
    {
        int []arr = { 2, 5, 3, 7, 11, 8, 10, 13, 6 };
        int n = arr.Length;

        Console.Write("LIS size\n" + 
                LongestIncreasingSubsequence(arr, n));
    }
}

// This code is contributed by nitin mittal.

Exercises:

1. You know Kadane‘s algorithm to find maximum sum sub-array. Modify Kadane’s algorithm to trace starting and ending location of maximum sum sub-array.

2. Modify Kadane‘s algorithm to find maximum sum sub-array in a circular array. Refer GFG forum for many comments on the question.

3. Given two integers A and B as input. Find number of Fibonacci numbers existing in between these two numbers (including A and B). For example, A = 3 and B = 18, there are 4 Fibonacci numbers in between {3, 5, 8, 13}. Do it in O(log K) time, where K is max(A, B). What is your observation?

— Venki. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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Improved By : nitin mittal




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